Question #299823

1. A sample of gaseous uranium hexafluoride, UF6, is held at a temperature of 300 K and a pressure of 0.1 mbar. The collision diameter of UF6 is 0.40 nm. (i) Calculate the root-mean-square (r.m.s.) speed of the molecules. Make sure to show all units in your working and how the units cancel out. [5 marks] (ii) Estimate the mean free path and collision frequency under these conditions. [10 marks]


1
Expert's answer
2022-02-20T08:36:17-0500

Vrms=3RTMV_{rms}=\sqrt{\frac{3RT}{M}}


Molar mass of UF6=352g/molUF_6=352g/mol

M=0.352kg/mol\therefore\>M=0. 352kg/mol




R=8.314J/mol/KR=8.314J/mol /K

T=300KT=300K



Vrms=3×8.314Jmol1K1300K0.352kgmol1V_{rms}=\frac{\sqrt{3×8.314Jmol^{-1}K^{-1}300K}}{0.352kgmol^{-1}}



=145.8Jkg1=145.8\sqrt{Jkg^{-1}}


But J=kgm2s2J=kgm^2s^{-2}

=145.8kgm2s2kg1=145.8\sqrt{kgm^2s^{-2}kg^{-1}}


=145.8m/s=145.8m/s


Part (ii)


Mean free path λ\lambda is given by


λ=KT2πd2P\lambda=\frac{KT}{\sqrt{2}\pi\>d^2P}


=1.38×1023×3002×π×(0.4×109)2×0.1×102=\frac{1.38×10^{-23}×300}{\sqrt{2}×\pi×(0.4×10^{-9})^2×0.1×10^2}


=5.824×104m=5.824×10^{-4}m


Collision frequency


=Vrmsλ=145.85.824×104=2.503×105s1=\frac{V_{rms}}{\lambda}=\frac{145.8}{5.824×10^{-4}}=2.503×10^5s^{-1}





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